1
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Question

# If 2x3+ax2+bxâˆ’6 has (xâˆ’1) as a factor and leaves a remainder 2 when divided by (xâˆ’2), find the value of 'a' and 'b' respectively.

A
8,12
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B
8,12
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C
4,10
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D
4,10
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Solution

## The correct option is A −8,12Let f(x)=2x3+ax2+bx−6Since (x−1) is the factor of f(x), therefore, by factor theorem, f(1)=0 that is:f(1)=2(1)3+a(1)2+(b×1)−6⇒0=(2×1)+(a×1)+b−6⇒0=2+a+b−6⇒a+b=6−2⇒a+b=4.......(1)Also, it is given that when f(x) is divided by (x−2), it leaves remainder 2, therefore, by remainder theorem, f(2)=2 that is:f(2)=2(2)3+a(2)2+(b×2)−6⇒2=(2×8)+(a×4)+2b−6⇒2=16+4a+2b−6⇒4a+2b=2+6−16⇒4a+2b=−8.......(2)Multiply equation 1 by 2 as follows:2(a+b)=2×4⇒2a+2b=8.....(3)Now, subtract equation 3 from equation 2 as shown below:(4a−2a)+(2b−2b)=−8−8⇒2a=−16⇒a=−162⇒a=−8Substituting the value of a in equation 1, we have:a+b=4⇒−8+b=4⇒b=4+8⇒b=12Hence, a=−8 and b=12.

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